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Dense subset of R

A set \(A\) is said to be a dense subset of \(R\) if \(\forall x,y\in R,x<y\), \(\exists a\in A\), such that \(x<a<y\). My this note is to show that the set \(S=\left\{ m+n\sqrt { 2 } :m,n\in Z \right\} \) is dense in \(R\). For that we observe that \(S\) satisfies the following properties:

i)Additivity: If \(x,y\in S\), then \(\left( x+y \right) \in S\).

ii)Homogeneity in \(Z\): If \(x\in S\) and \(k\in Z\), then \(kx\in S\)

Now let \(\varepsilon >0\) be a real to be chosen such that \(b-a>\varepsilon ,a,b\in R\) be given. Then it is trivial to show that \(\exists {n}_{1}\in Z\) such that \(a<{n}_{1}\varepsilon <b\). Now we take \(\varepsilon =\frac { b-a }{ 2 } \). Then \(\exists s\in S\) such that \(0<s<\varepsilon \). Hence, \(\exists {n}_{2}\in Z\) such that \(a<{n}_{2}s<b\). Since by property ii), \({n}_{2}s\in S\),the property is proved.

There is another way to prove the above proposition. First prove that every proper subgroup of the additive group \(R\) is either dense or cyclic in \(R\). Then prove that the given subset \(S\) of \(R\) is a subgroup of \(R\) that is not cyclic in \(R\)..hence it is dense in \(R\).